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# Find the equation of parabola with vertex (2,-3) and directrix x-4y-48=0.

Toolbox:
• Directrix of a parabola is a line $\perp$ to axis of the parabola.
• Vertex (A), focus (S) and foot of $\perp$ from vertex on the directrix (A') all lie on the axis of the parabola.
• As per the def. of parabola Vertex is the mid point of focus and foot of $\perp$ of vertex on directrix.
Given: vertex $A(2,-3).$ and directrix $x-4y-48=0$.............$(i)$
We know that Axis of parabola is $\perp$ to directrix and
it passes through focus and vertex of the parabola.
Equation of a line $\perp$ to the line $ax+by+c=0$ is given by $bx-ay+c_1=0$.
$\therefore$ The equation of the axis is $-4x-y+c=0$
or $4x+y+c=0$
But vertex $(2,-3)$ lies on the axis of the parabola.
$\therefore\:$It satisfies the equation of the axis. $\Rightarrow\:8-3+c=0$ or $c=-5$
$\therefore$ The equation of axis is $4x+y-5=0$..............$(ii)$
Foot of $\perp$ from vertex to directrix(A') is got by solving the equations of axis and directrix.
By solving $(i)\:\:and\:\:(ii)$ we get the foot of $\perp$ of vertex $(A')$ on directrix which is
$(x-4y-48)+4(4x+y-5)=0$
$\Rightarrow\:17x-68=0$ or $x=4\:\:and\:\:y=-11$
$\therefore A'(4,-11)$
We know that Vertex(A) is the mid point of focus(S) and foot of $\perp$ $(A')$
Let focus be $S(x,y)$
$\therefore \:\big(\large\frac{x+4}{2},\frac{y-11}{2}\big)=(2,-3)$
$\Rightarrow\:x=0\:\:and\:\:y=5$
$i.e., focus\: (S)=(0,5)$
We know that equation of the parabola is equation of locus of a point which moves
such that its distance from focus = its distance from directrix.
$\therefore$ Equation of the parabola is:
Let $P(x,y)$ be any point on the parabola.
As per the def. of parabola Distance PS=Distance P$\rightarrow\:Directrix (i)$
$\Rightarrow\:\sqrt{(x-0)^2+(y-5)^2}=\large\frac{x-4y-48}{\sqrt {1+16}}$
Squaring on both the sides we get
$17x^2+17y^2+425-170y=x^2+16y^2+48^2-8xy+384y-96x$
$\Rightarrow\: 16x^2+y^2+8xy+96x-554y-1879=0$